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A car dealership sells a particular luxury car at a rate of 1.1 per week. What is the probability that it will sell three cars next week?

User Varna
by
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2 Answers

3 votes

Answer:

0.0733, or 7.33%.

Explanation:

The answer above is correct.

User Poshanniraula
by
7.4k points
3 votes

Answer:

The probability of selling three cars next week is approximately 0.0733, or 7.33%.

Explanation:

We can use the Poisson distribution to solve this problem, since it models the number of events (in this case, the number of cars sold) that occur in a fixed period of time, given a known rate of occurrence.

The Poisson distribution has the following formula:

P(x) = (e^(-λ) * λ^x) / x!

where:

  • P(x) is the probability of x events occurring
  • e is the mathematical constant (approximately equal to 2.71828)
  • λ is the average rate of occurrence
  • x is the number of events that we want to calculate the probability for
  • x! is the factorial of x (i.e., x! = x * (x-1) * (x-2) * ... * 2 * 1)

In this case, we know that the rate of occurrence (λ) is 1.1 cars per week, and we want to calculate the probability of selling 3 cars (x = 3) next week. Substituting these values into the Poisson distribution formula, we get:

P(3) = (e^(-1.1) * 1.1^3) / 3!

P(3) = (0.332871 * 1.331) / 6

P(3) = 0.0733

Therefore, the probability of selling three cars next week is approximately 0.0733, or 7.33%.

User Gonso
by
7.0k points